An intro to D-modules
Daniel Smolkin
BIKES, February 9, 2016
1
The Weyl algebra
Let K[x1, · · · , xn] (henceforth abbreviated K[X]) be a polynomial ring. The nth Weyl
algebra is a K-subalgebra of EndK(K[X]) generated by:
• ˆxi, where ˆxi(g) = xig
• ∂ ∂xi
Typically we denote ˆxi simply by xi and ∂x∂
i by ∂i. So a typical element of A2 might look
like σ = ∂12∂2x1x2− 4x31x 2 2 where σ(x21) = ∂ 3 ∂2x 1∂x2 x1x2x21− 4x 3 1x 2 2x 2 1= ∂3 ∂2x 1∂x2 x31x2− 4x51x 2 2= 6x1− 4x51x 2 2
We denote this Weyl algebra An. Note that the Weyl algebra is NOT commutative! It
comes with a commutator [·, ·] defined by [a, b] = ab − ba. Here are some commutators to know:
• [∂i, xi] = 1, because for any f ∈ K[X], the product rule says
[∂i, xi]f = (∂ixi− xi∂i) f = ∂i(xif ) − xi(∂if ) = f + xi∂if − xi∂if = 1 · f
• More generally, [∂i, f ] = ∂x∂f
i (this is an element of the Weyl algebra!)
• Thus, [∂i, xj] = δij
• [xi, xj] = 0 for all i, j
1.1
Canonical form
Writing xα1
1 · · · xαnn is clunky, so we adopt the following notation: given α = (α1, · · · , αn) ∈
Nn, by xα we mean xα11· · · xαnn. Similarly for ∂α. Such an α is called a multi-index. Also,
by |α| we mean α1+ · · · αn. If char k = 0, The Weyl algebra An has K-vector space basis
given byxα∂β | α, β ∈ Nn . For instance,
∂i2x2i = ∂i(xi2∂i+ 2xi) = ∂ix2i∂i+ 2∂ixi= (x2i∂i+ 2xi)∂i+ 2xi∂i+ 2
What goes wrong if char k = p? Well, in that case ∂pi = 0! Indeed, if a ≥ p, then ∂ipxai = 0.
This canonical form is our friend and most proofs will rely on looking at it and applying some commutators cleverly.
1.2
Degree of an operator
Given a monomial xα∂β ∈ A
n, we can define degxα∂β = |α| + |β|. Given any element of
An, we define its degree to be the largest degree among the degrees of its monomials. It
takes a bit of work, but one can show Proposition 1.1. We have the following:
• deg(D + D0) ≤ max{deg D, deg D0}
• deg(DD0) = deg D + deg D0
• deg[D, D0] ≤ deg D + deg D0− 2
Proof. 1 is clear. For 2 and 3, Induce on degD + degD0. Also, from 1, it’s enough to assume that D and D0 are monomials.
This shows that An is a domain. Also, we have the following neat fact:
Proposition 1.2. An has no nontrivial two-sided ideals (so it’s a simple algebra)
Proof. Proof by contradiction. Suppose I is a nonzero two-sided ideal. Choose D of minimal degree k in I. Then [D, f ] ∈ I for all f . Now, deg D > 0, so there is some α and β such that xα∂β has a nonzero coefficient in the expansion of D, say β
i > 0. But then [xi, D] is
nonzero and has degree ≤ k − 1. Thus, β = 0. But if αi> 0, then [D, ∂i] has degree ≤ k − 1
and is not zero.
2
Modules over A
nWe haven’t even defined a D-module yet! A D-module is a module over An (or the ring of
differential operators of any ring).
3
Graded and filtered modules
Recall that a ring R is graded if R =L
iRiwith Ri· Rj ⊆ Ri+j. If R is a graded ring, then
a graded R-module is a module M such that M =L
iMi where RiMj⊆ Mi+j.
Now, we’d like to make Aninto a graded algebra. There’s one problem though: we can’t
say something like “∂1x1 degree 2” because ∂1x1 = x1∂1+ 1. So instead we have to be
content with a filtration. Recall that a filtration of a K-algebra R is an ascending chain of K-vector spaces F0⊂ F1⊂ F2. . . such that FiFj⊂ Fi+j.
One filtration on An that’s of particular interest to us is the Bernstein filtration, denoted
Bi. We define Bito be the k-vectorspace generated byxα∂β| |α| + |β| ≤ i . This filtration
is cool because every Bi is a finite-dimensional vector space over k.
One more construction: let R be a ring and {Fi} a filtration of R. For all n, let σn :
Fn → Fn/Fn−1 be the usual quotient map. This map σ is called the symbol map in this
context. Then the graded ring associated to F , denoted grF(R), is given by
grF(R) =M
i
Fi/Fi−1
Note that any homogenous element of degree k in this ring can be written as σk(q) for some
q ∈ Fk.
Lemma 3.1. Sn is a polynomial ring in 2n variables.
Proof. It’s easy to see that grβA
nis generated by σ1(xi) and σ1(δi) over k. Let yi= σ(xi) ∈
Sn and yn+i= ∂i∈ Sn. We wish to show that the yicommute. For this it’s enough to show
that yiyn+i= yn+iyi. But this is clear: σ2(∂ixi) = σ2(xi∂i+ 1) = σ2(x2∂i).
Thus we can write a surjective map from a polynomial ring k[z1, · · · , z2n] → Sn by
zj 7→ yj, for 1 ≤ j ≤ 2n. It remains to show that this map, ϕ, is injective. To that end,
suppose some homogenous polynomial F is sent to zero. If ϕ(F ) = 0, then ϕ(F ) = σk(d)
where d = F (x1, · · · , xn, ∂1, · · · , ∂n). But if σk(d) = 0, then d can also be written as a sum
of monomomials of degree less than k. So d must be zero to begin with, since canonical form is a k-basis of An.
Similarly define a filtered module and graded module associated to a filtration: if M is an An module, then a filtration of M with respect to B is an ascending chain of k-vector
spaces Γ0⊆ Γ1⊆ · · · with M =SiΓi and BiΓi⊆ Γi+j.
A filtration on M induces filtrations on its submodules and quotients: let Γ be a filtration on M with respect to B and let N ⊆ M be a submodule. Then Γi∩ N and Γi+ N/N are
filtrations on N and M/N , repesctively, that agree with B. Now, let M be a left An module. Then
Lemma 3.2. if grΓM is a (left) noetherian S
n module, then M is a noetherian An module
Proof. Pick any N ⊆ M and let Γ0 be the induced grading on N . Then grΓ0N ⊆ grΓM is a
finitely generated submodule. We wish to show that N is a finitely generated submodule of M .
Since grΓ0N is finitely generated, we can enumerate its generators f
1, · · · , fs. Then
there is some m such that grΓ0N is generated by elements of degree ≤ m. I claim that
N is generated by elements of degree ≤ m. Suppose this weren’t the case: then choose a homogenous element f ∈ N of minimal degree such that f 6∈ An · Γ0m. Then f has
some degree k > m. Then σk(f ) = P σk−ri(ai)fi for some ai ∈ Sn. But this implies
f −P aifˆi∈ ker σk= Γk−1, where fi= µ( ˆfi). This contradicts the fact that deg f = k.
Thus N is generated by elements in Γ0
m. But Γ0m is a finite-dimensional k-vector space
with a finite dimensional basis. So this basis generates N .
The converse doesn’t hold. If grΓM is noetherian, we call Γ a good filtration of M .
Theres a lemma saying all good filtrations are sorta the same. Namely:
Lemma 3.3. A filtration Γ of M is good if and only if there exists some k0 such that
Γi+k= Biγk for all k > k0.
and
Lemma 3.4. Let Γ and Ω be two filtrations of M . If Γ is good, then there is some k1 such
that Γj⊆ Ωj+k1 for all j.
Proof. This follows from the above: choose some k0 such that Γi+k0 = BiΓk0. Then there
is some k1 such that Ωk1 ⊇ Γk0, since Γk0 is a finite-dimensional k-vector space. Then for
any j,
Note that any Noetherian Anmodule has a good filtration: if M is generated by u1, · · · uk,
set Γj=P k i=1Bjui.
4
Dimension and multiplicity
Here’s a result from commutative algebra: Theorem 4.1. Let M =L
i≥0Mibe a graded module over a polynomial ring K[x1, · · · , xn].
There exists a polynomial h(t) ∈ Q[t] such that
s
X
i=0
dimkMi= h(s)
for s >> 0.
Let M be a module over An and let h be the hilbert polynomial of grΓM , where Γ is a
good filtration of M with respect to An. We define the dimension of M to be the degree d
of h, and we define the multiplicity of M to be d!ad, where ad is the leading coefficient. It’s
not hard to see this is indepedent of good filtration using 4.3. Example: d(K[x1, · · · , xn] = n and d(An) = 2n.
These have some nice properties:
Lemma 4.2. Let M be a finitely generated An module and N a submodule. Then
• dim M = max {dim N, dim M/N }
• More generally, if M = M1⊕ · · · ⊕ Ms, then dim M = max{dim M1, · · · , dim Ms}
• m(M ) = m(N ) + m(M/N ) if dim N = dim M/N
Proof. Let Γ be the good filtration of M . Let Γ0 and Γ00 be the induced filtrations on N and M/N , respectively. It’s easy to see that
0 → grΓ0N → grΓM → grΓ00M/N → 0
whence Γ0 and Γ00 are good filtrations. This allows us to see hN+ hM/N = hM. The lemma
follows.
Now let M be a finitely generated An module. Since we have a surjection An→ M , the
above lemma tells us that dim M ≤ dim An= 2n. Surprisingly, we get another inequality:
Theorem 4.3 (Bernstein’s inequality). If M is a finitely-generated An-module, n ≤ dim M ≤
2n
Proof. First, we define a map Bi→ Homk(Γi, Γ2i) where b get’s sent to “multiplication by
b”. This is an injective map. The proof uses induction and some manipulation of canonical forms. Indeed, to see this, it’s enough to show that aΓi 6= 0 for any 0 6= a ∈ Bi. We proceed
by induction. The base case is easy since B0= k.
Now, if aΓi = 0, then a 6∈ K, and hence the canonical form of a has some term cxα∂β
where c 6= 0, and |α| + |β| > 0. Then [a, ∂i] is not zero and it’s in Bi. Now,
but aΓi−1= 0 because aΓi= 0 and Γi contains Γi−1. Also, ∂iΓi−1⊆ Γi So,
[a, ∂i]Γi−1= 0
but this contradicts the induction hypothesis. We get that dim Bi ≤ dim Hom(Γi, Γ2i) =
χ(i)χ(2i). So χ(i)χ(2i) is a polynomial of degree as least 2n in i. But, the degree of this polynomial is also 2d(M ), so d(M ) ≥ n.
5
Holonomic modules
. . . are D-modules of minimal dimension.
It’s easy to see holonomic modules are artinian, using multiplicity! Indeed, the length of a holonomic D-module cannot exceed its length.
6
Differential operators (time permitting)
Let R be a commutative k-algebra. Then we define the differential operators of order ≤ n as follows:
• the differential operators of order 0 are the elements of End R whose commutator with anythign in R is zero
• order ≤ n if [a, P ] has order ≤ n − 1 for all a ∈ R. operators of order 1 turn out to just be the derivations.
The Weyl algebra is D(k[x1, · · · , xn])
6.1
Other construction of the Weyl algebra
It turns out the commutator relations above characterize the Weyl algebra. To be more precise, let R = K {z1, · · · , z2n} be the free (non-commutative!) K-algebra in 2n generators.
Then we get a surjective map R → An given by zi7→ xifor i ≤ n, and zn+j7→ ∂j for j ≤ n.
If char k = 0, the kernel of this map is exactly the two-sided ideal generated by • [zi, zj], |i − j| 6= n
• [zn+i, zi] = 1
Otherwise, the kernel includes zpn+i for all i. So there are two possible ways to define the Weyl algebra in characteristic p! I think the preference is to use something called “divided powers”. . . or just the ring of differential operators!
